IRU3138CS datasheet(10/18 Pages) IRF

10 / 18 page

Rev. 1.0

01/29/04

IRU3138

www.irf.com

The IRU3138’s error amplifier is a differential-input

transconductance amplifier. The output is available for

DC gain control or AC phase compensation.

The E/A can be compensated with or without the use of

local feedback. When operated without local feedback,

the transconductance properties of the E/A become evi-

dent and can be used to cancel one of the output filter

poles. This will be accomplished with a series RC circuit

from Comp pin to ground as shown in Figure 9.

Note that this method requires that the output capacitor

should have enough ESR to satisfy stability requirements.

In general, the output capacitor’s ESR generates a zero

typically at 5KHz to 50KHz which is essential for an

acceptable phase margin.

The ESR zero of the output capacitor expressed as fol-

lows:

Figure 9 - Compensation network without local

feedback and its asymptotic gain plot.

The transfer function (Ve / VOUT) is given by:

The (s) indicates that the transfer function varies as a

function of frequency. This configuration introduces a gain

and zero, expressed by:

|H(s)| is the gain at zero cross frequency.

First select the desired zero-crossover frequency (Fo):

Use the following equation to calculate R4:

To cancel one of the LC filter poles, place the zero be-

fore the LC filter resonant frequency pole:

Using equations (17) and (19) to calculate C9, we get:

One more capacitor is sometimes added in parallel with

C9 and R4. This introduces one more pole which is mainly

used to suppress the switching noise. The additional

pole is given by:

The pole sets to one half of switching frequency which

results in the capacitor CPOLE:

≅ 2.4nF; Choose C9=2.2nF

FESR =

---(14)

p3ESR3Co

H(s) =

---(15)

(

)

R6 + R5

1 + sR4C9

sC9

R4 =

---(18)

3FESR

FLC2

VOSC

VIN

R5 + R6

Fo > FESR and FO

[ (1/5 ~ 1/10)3fS

For:

Lo = 1.1

Co = 990

≅ 75%FLC

≅ 0.753

3 CO

---(19)

FZ = 3.6KHz

R4 = 17.8K

FP =

p3R43

3CPOLE

C9 + CPOLE

FZ =

---(17)

p3R43C9

|H(s=j

32p3FO)| = gm3

3R4

---(16)

3R5

VOUT

Vp=VREF

E/A

H(s) dB

Frequency

Gain(dB)

Comp

Optional

FLC = 4.82KHz

R5 = 1K

R6 = 1K

gm = 600

mmho

For:

VIN = 5V

VOSC = 1.25V

Fo = 40KHz

FESR = 12KHz

This results to R4=17.32K

Choose R4=17.8K

Where:

VIN = Maximum Input Voltage

VOSC = Oscillator Ramp Voltage

Fo = Crossover Frequency

FESR = Zero Frequency of the Output Capacitor

FLC = Resonant Frequency of the Output Filter

R5 and R6 = Resistor Dividers for Output Voltage

Programming

gm = Error Amplifier Transconductance

CPOLE =

≅

p3R43fS

p3R43fS -

For FP << fS/2

R4=17.8K and FS=400KHz will result to CPOLE=44pF.

Choose CPOLE=47pF.

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